Does the triangle inequality follow from the rest of the properties of a subfield-valued absolute value?

Question

(This is a much more specific version of my earlier question from over a year ago.)

Let $F$ be a field, let $E$ be an ordered subfield of $F$, and let $\;\; |\hspace{-0.03 in}\cdot\hspace{-0.03 in}| \: : \: F \: \to \: E$
be such that for all members $x$ of $E$, $\:$ for all members $y$ and $z$ of $F$,

$(1)$ $0\leq x \: \implies \: |\hspace{.01 in}x\hspace{.01 in}| = x$

$(2)$ $0\leq |y\hspace{.01 in}|$

$(3)$ $|y\cdot z\hspace{.01 in}| \: = \: |y\hspace{.01 in}| \cdot |\hspace{.01 in}z\hspace{.01 in}|$

Does it follow that for all members $y$ and $z$ of $\hspace{.01 in}F$, $\;\; |y+z\hspace{.01 in}| \: \leq \: |y\hspace{.01 in}| + |\hspace{.01 in}z\hspace{.01 in}| \;\;\;\;$?

I suspect the answer is no, but I have not been able to come up with any counterexample.

Answer 1

Let $F=\mathbb Q((X))$ be the field of formal Laurent series with rational coefficients, $E=\mathbb Q$. If $a=\sum_{k=n}^\infty a_k X^k$ with $a_n\ne 0$, let $|a|=|a_n|$ (and let $|0|=0$). Then $(1),(2),(3)$ are readily verified. But $|1+2X|= 1$, $|-1+2X|=1$, $|(1+2X)+(-1+2X)|=|4X|=4>1+1$.